Calculus Examples

$f (x) = 3 x^{5} - 5 x^{3}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

By the Sum Rule, the derivative of $3 x^{5} - 5 x^{3}$ with respect to $x$ is $\frac{d}{d x} [3 x^{5}] + \frac{d}{d x} [- 5 x^{3}]$ .

$f' (x) = \frac{d}{d x} (3 x^{5}) + \frac{d}{d x} (- 5 x^{3})$

Step 1.1.1.2

Evaluate $\frac{d}{d x} [3 x^{5}]$ .

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Step 1.1.1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{5}$ with respect to $x$ is $3 \frac{d}{d x} [x^{5}]$ .

$f' (x) = 3 \frac{d}{d x} (x^{5}) + \frac{d}{d x} (- 5 x^{3})$

Step 1.1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f' (x) = 3 (5 x^{4}) + \frac{d}{d x} (- 5 x^{3})$

Step 1.1.1.2.3

Multiply $5$ by $3$ .

$f' (x) = 15 x^{4} + \frac{d}{d x} (- 5 x^{3})$

Step 1.1.1.3

Evaluate $\frac{d}{d x} [- 5 x^{3}]$ .

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Step 1.1.1.3.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{3}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 15 x^{4} - 5 \frac{d}{d x} x^{3}$

Step 1.1.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 15 x^{4} - 5 (3 x^{2})$

Step 1.1.1.3.3

Multiply $3$ by $- 5$ .

$f' (x) = 15 x^{4} - 15 x^{2}$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

By the Sum Rule, the derivative of $15 x^{4} - 15 x^{2}$ with respect to $x$ is $\frac{d}{d x} [15 x^{4}] + \frac{d}{d x} [- 15 x^{2}]$ .

$f'' (x) = \frac{d}{d x} (15 x^{4}) + \frac{d}{d x} (- 15 x^{2})$

Step 1.1.2.2

Evaluate $\frac{d}{d x} [15 x^{4}]$ .

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Step 1.1.2.2.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x^{4}$ with respect to $x$ is $15 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 15 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 15 x^{2})$

Step 1.1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 15 (4 x^{3}) + \frac{d}{d x} (- 15 x^{2})$

Step 1.1.2.2.3

Multiply $4$ by $15$ .

$f'' (x) = 60 x^{3} + \frac{d}{d x} (- 15 x^{2})$

Step 1.1.2.3

Evaluate $\frac{d}{d x} [- 15 x^{2}]$ .

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Step 1.1.2.3.1

Since $- 15$ is constant with respect to $x$ , the derivative of $- 15 x^{2}$ with respect to $x$ is $- 15 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 60 x^{3} - 15 \frac{d}{d x} x^{2}$

Step 1.1.2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 60 x^{3} - 15 (2 x)$

Step 1.1.2.3.3

Multiply $2$ by $- 15$ .

$f'' (x) = 60 x^{3} - 30 x$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $60 x^{3} - 30 x$ .

$60 x^{3} - 30 x$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $60 x^{3} - 30 x = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$60 x^{3} - 30 x = 0$

Step 1.2.2

Factor $30 x$ out of $60 x^{3} - 30 x$ .

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Step 1.2.2.1

Factor $30 x$ out of $60 x^{3}$ .

$30 x (2 x^{2}) - 30 x = 0$

Step 1.2.2.2

Factor $30 x$ out of $- 30 x$ .

$30 x (2 x^{2}) + 30 x (- 1) = 0$

Step 1.2.2.3

Factor $30 x$ out of $30 x (2 x^{2}) + 30 x (- 1)$ .

$30 x (2 x^{2} - 1) = 0$

Step 1.2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$2 x^{2} - 1 = 0$

Step 1.2.4

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.5

Set $2 x^{2} - 1$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $2 x^{2} - 1$ equal to $0$ .

$2 x^{2} - 1 = 0$

Step 1.2.5.2

Solve $2 x^{2} - 1 = 0$ for $x$ .

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Step 1.2.5.2.1

Add $1$ to both sides of the equation.

$2 x^{2} = 1$

Step 1.2.5.2.2

Divide each term in $2 x^{2} = 1$ by $2$ and simplify.

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Step 1.2.5.2.2.1

Divide each term in $2 x^{2} = 1$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{1}{2}$

Step 1.2.5.2.2.2

Simplify the left side.

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Step 1.2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{1}{2}$

Step 1.2.5.2.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{1}{2}$

Step 1.2.5.2.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{\frac{1}{2}}$

Step 1.2.5.2.4

Simplify $\pm \sqrt{\frac{1}{2}}$ .

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Step 1.2.5.2.4.1

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{2}}$

Step 1.2.5.2.4.2

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{2}}$

Step 1.2.5.2.4.3

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Step 1.2.5.2.4.4

Combine and simplify the denominator.

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Step 1.2.5.2.4.4.1

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Step 1.2.5.2.4.4.2

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Step 1.2.5.2.4.4.3

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Step 1.2.5.2.4.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Step 1.2.5.2.4.4.5

Add $1$ and $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

Step 1.2.5.2.4.4.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 1.2.5.2.4.4.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$x = \pm \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Step 1.2.5.2.4.4.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \pm \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Step 1.2.5.2.4.4.6.3

Combine $\frac{1}{2}$ and $2$ .

$x = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 1.2.5.2.4.4.6.4

Cancel the common factor of $2$ .

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Step 1.2.5.2.4.4.6.4.1

Cancel the common factor.

$x = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 1.2.5.2.4.4.6.4.2

Rewrite the expression.

$x = \pm \frac{\sqrt{2}}{2^{1}}$

Step 1.2.5.2.4.4.6.5

Evaluate the exponent.

$x = \pm \frac{\sqrt{2}}{2}$

Step 1.2.5.2.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.5.2.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{\sqrt{2}}{2}$

Step 1.2.5.2.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \frac{\sqrt{2}}{2}$

Step 1.2.5.2.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Step 1.2.6

The final solution is all the values that make $30 x (2 x^{2} - 1) = 0$ true.

$x = 0, \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, - \frac{\sqrt{2}}{2}) \cup (- \frac{\sqrt{2}}{2}, 0) \cup (0, \frac{\sqrt{2}}{2}) \cup (\frac{\sqrt{2}}{2}, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - \frac{\sqrt{2}}{2})$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 3$ in the expression.

$f'' (- 3) = 60 {(- 3)}^{3} - 30 \cdot - 3$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raise $- 3$ to the power of $3$ .

$f'' (- 3) = 60 \cdot - 27 - 30 \cdot - 3$

Step 4.2.1.2

Multiply $60$ by $- 27$ .

$f'' (- 3) = - 1620 - 30 \cdot - 3$

Step 4.2.1.3

Multiply $- 30$ by $- 3$ .

$f'' (- 3) = - 1620 + 90$

Step 4.2.2

Add $- 1620$ and $90$ .

$f'' (- 3) = - 1530$

Step 4.2.3

The final answer is $- 1530$ .

$- 1530$

Step 4.3

The graph is concave down on the interval $(- \infty, - \frac{\sqrt{2}}{2})$ because $f'' (- 3)$ is negative.

Concave down on $(- \infty, - \frac{\sqrt{2}}{2})$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- \frac{\sqrt{2}}{2}, 0)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $- 0.35$ in the expression.

$f'' (- 0.35) = 60 {(- 0.35)}^{3} - 30 \cdot - 0.35$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Raise $- 0.35$ to the power of $3$ .

$f'' (- 0.35) = 60 \cdot - 0.042875 - 30 \cdot - 0.35$

Step 5.2.1.2

Multiply $60$ by $- 0.042875$ .

$f'' (- 0.35) = - 2.5725 - 30 \cdot - 0.35$

Step 5.2.1.3

Multiply $- 30$ by $- 0.35$ .

$f'' (- 0.35) = - 2.5725 + 10.5$

Step 5.2.2

Add $- 2.5725$ and $10.5$ .

$f'' (- 0.35) = 7.9275$

Step 5.2.3

The final answer is $7.9275$ .

$7.9275$

Step 5.3

The graph is concave up on the interval $(- \frac{\sqrt{2}}{2}, 0)$ because $f'' (- 0.35)$ is positive.

Concave up on $(- \frac{\sqrt{2}}{2}, 0)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(0, \frac{\sqrt{2}}{2})$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $0.35$ in the expression.

$f'' (0.35) = 60 {(0.35)}^{3} - 30 \cdot 0.35$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Raise $0.35$ to the power of $3$ .

$f'' (0.35) = 60 \cdot 0.042875 - 30 \cdot 0.35$

Step 6.2.1.2

Multiply $60$ by $0.042875$ .

$f'' (0.35) = 2.5725 - 30 \cdot 0.35$

Step 6.2.1.3

Multiply $- 30$ by $0.35$ .

$f'' (0.35) = 2.5725 - 10.5$

Step 6.2.2

Subtract $10.5$ from $2.5725$ .

$f'' (0.35) = - 7.9275$

Step 6.2.3

The final answer is $- 7.9275$ .

$- 7.9275$

Step 6.3

The graph is concave down on the interval $(0, \frac{\sqrt{2}}{2})$ because $f'' (0.35)$ is negative.

Concave down on $(0, \frac{\sqrt{2}}{2})$ since $f'' (x)$ is negative

Step 7

Substitute any number from the interval $(\frac{\sqrt{2}}{2}, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 7.1

Replace the variable $x$ with $3$ in the expression.

$f'' (3) = 60 {(3)}^{3} - 30 \cdot 3$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $3$ to the power of $3$ .

$f'' (3) = 60 \cdot 27 - 30 \cdot 3$

Step 7.2.1.2

Multiply $60$ by $27$ .

$f'' (3) = 1620 - 30 \cdot 3$

Step 7.2.1.3

Multiply $- 30$ by $3$ .

$f'' (3) = 1620 - 90$

Step 7.2.2

Subtract $90$ from $1620$ .

$f'' (3) = 1530$

Step 7.2.3

The final answer is $1530$ .

$1530$

Step 7.3

The graph is concave up on the interval $(\frac{\sqrt{2}}{2}, \infty)$ because $f'' (3)$ is positive.

Concave up on $(\frac{\sqrt{2}}{2}, \infty)$ since $f'' (x)$ is positive

Step 8

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(- \infty, - \frac{\sqrt{2}}{2})$ since $f'' (x)$ is negative

Concave up on $(- \frac{\sqrt{2}}{2}, 0)$ since $f'' (x)$ is positive

Concave down on $(0, \frac{\sqrt{2}}{2})$ since $f'' (x)$ is negative

Concave up on $(\frac{\sqrt{2}}{2}, \infty)$ since $f'' (x)$ is positive

Step 9